Problem: Let $g(x)=\begin{cases} 5-x&\text{for }x \leq 3 \\\\ 2e^{3-x}&\text{for }x>3 \end{cases}$ Is $g$ continuous at $x=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Explanation: For $g$ to be continuous at $x=3$, we need $\lim_{x\to 3}g(x)$ and $g(3)$ to exist and be equal. Since $3\leq 3$, the rule that applies to $x=3$ is $5-x$. So $g(3)=5-3=2$. Now let's analyze $\lim_{x\to 3}g(x)$. Finding $\lim_{x\to 3^{ +}}g(x)$ For $x$ -values larger than $3$, the appropriate rule for $g(x)$ is $2e^{3-x}$. Since $2e^{3-x}$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 3^{ +}}g(x) \\\\ &=\lim_{x\to 3^{ +}}[2e^{3-x}] \gray{2e^{3-x}\text{ is the rule for }x>3} \\\\ &=2e^{3-3} \gray{2e^{3-x}\text{ is continuous at }x=3} \\\\ &=2 \end{aligned}$ Finding $\lim_{x\to 3^{ -}}g(x)$ For $x$ -values smaller than $3$, the appropriate rule for $g(x)$ is $5-x$. Since $5-x$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 3^{ -}}g(x) \\\\ &=\lim_{x\to 3^{ -}}[5-x] \gray{5-x\text{ is the rule for }x<3} \\\\ &=5-3 \gray{5-x\text{ is continuous at }x=3} \\\\ &=2 \end{aligned}$ Conclusion We found that: $\lim_{x\to 3^{ +}}g(x)=\lim_{x\to 3^{ -}}g(x)=g(3)=2$ Since the one-sided limits are both equal to $g(3)$, we can determine that the two-sided limit $\lim_{x\to 3}g(x)$ is also equal to $g(3)$, and $g$ is continuous at $x=3$.